# When 8.05 g of an unknown compound X was dissolved in 100. g of benzene (C6H6), the vapor pressure of the benzene decreased from 100.0 torr to 94.8 torr at 26 degrees C. What is the mole fraction and molar mass of X?

##### 1 Answer
Feb 23, 2017

I got $\text{114.51 g/mol}$.

The reduction of vapor pressure is a colligative property. From Raoult's law, we have:

${P}_{j} = {\chi}_{j} {P}_{j}^{\text{*}}$,

where ${P}_{j}$ is the partial pressure above the solution from component $j$, ${P}_{j}^{\text{*}}$ is the pure vapor pressure of $j$ by itself, and ${\chi}_{j}$ is the mol fraction of $j$ in the solution.

(We consider $\boldsymbol{j}$ to be the solvent and $\boldsymbol{i}$ to be the solute.)

But mol fractions are always $\le 1$. Therefore, for a two-component solution, ${\chi}_{j} = 1 - {\chi}_{i}$. Thus:

${P}_{j} = \left(1 - {\chi}_{i}\right) {P}_{j}^{\text{*}}$

Since ${\chi}_{j} \le 1$, it follows that $1 - {\chi}_{i} \le 1$ as well, or that ${\chi}_{i} \ge 0$. Therefore, adding any solute into a solvent decreases its vapor pressure.

This reduction is calculated as follows. Define $\Delta {P}_{j} = {P}_{j} - {P}_{j}^{\text{*}}$ to get:

$\textcolor{g r e e n}{\Delta {P}_{j}} = {\chi}_{j} {P}_{j}^{\text{*" - P_j^"*}}$

$= \left({\chi}_{j} - 1\right) {P}_{j}^{\text{*}}$

$= \left(1 - {\chi}_{i} - 1\right) {P}_{j}^{\text{*}}$

$= \textcolor{g r e e n}{- {\chi}_{i} {P}_{j}^{\text{*}}}$

Since ${P}_{j}^{\text{*}}$ and ${\chi}_{i}$ are positive, the reduction in vapor pressure is of course negatively-signed. From this equation then, we can determine the mol fraction of what was dissolved in benzene.

Plug ${P}_{j} = \text{94.8 torr}$ and ${P}_{j}^{\text{*" = "100.00 torr}}$ in to get:

(94.8 - 100.00) "torr" = -chi_i ("100.00 torr")

$\implies \textcolor{b l u e}{{\chi}_{i} = 0.052}$

Now that we have the mol fraction of $X$, we can find the molar mass of $X$ simply by solving for the mols of $X$, and dividing the given mass by the mols of $X$.

$0.052 = \frac{{n}_{X}}{{n}_{X} + {n}_{\text{Ben}}}$

$\implies 0.052 {n}_{X} + 0.052 {n}_{\text{Ben}} = {n}_{X}$

$\implies \left(0.052 - 1\right) {n}_{X} = - 0.052 {n}_{\text{Ben}}$

$\implies {n}_{X} = 0.0549 {n}_{\text{Ben}}$

But we know the molar mass of benzene; it's $6 \times 12.011 + 6 \times 1.0079 = \text{78.1134 g/mol}$. So, we get the mols of benzene by its molar mass:

$\text{100.0 g Benzene" xx "1 mol"/"78.1134 g" = "1.2802 mols Benzene solvent}$

Therefore, the mols of $X$ are:

n_X = 0.0549("1.2802 mols") = "0.0703 mols X"

This gives a molar mass for $X$ as:

color(blue)(M_(m,X)) = "8.05 g X"/"0.0703 mols X" = color(blue)("114.51 g/mol")